Question

Factorise : $27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$

Answer 1

$27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p=(3p{)}^3+{(-\frac{1}{6})}^3+3(3p\left)(-\frac{1}{6})(3p-\frac{1}{6})Usingidentity\right(a-b{)}^3=a^3-b^3-3ab(a-b)={(3p-\frac{1}{6})}^3=(3p-\frac{1}{6})(3p-\frac{1}{6})(3p-\frac{1}{6})$